= Solution
Use the same length, thermal-time, velocity and temperature scales as in thermal convection. Write the imposed <magnetic field> as $B_0\hat{\mathbf z}$ and measure its perturbation in units of $B_0$. The physical linear <magnetohydrodynamic momentum equation> contains magnetic tension $(B_0^2/\mu_0)\partial_z\mathbf b$; the gradient part of the <Lorentz force density> is absorbed into <magnetohydrodynamic total pressure>.
Divide the momentum equation by the viscous force scale $\rho_0\nu\kappa/d^3$. Its acceleration coefficient is $\kappa/\nu=1/\sigma$ and its magnetic coefficient is
$$
\frac{B_0^2d^2}{\mu_0\rho_0\nu\kappa}=Q\zeta,\qquad \zeta=\frac\eta\kappa,\qquad Q=\frac{B_0^2d^2}{\mu_0\rho_0\eta\nu}.
$$
Here $Q$ is the <Chandrasekhar number>, comparing magnetic and viscous-diffusive effects, and $\zeta$ is the ratio of <magnetic diffusivity> to <thermal diffusivity>. The induction scale is $B_0\kappa/d^2$, so linearization gives
$$
\boxed{\frac1\sigma\dot{\mathbf u}=-\nabla P+Q\zeta\partial_z\mathbf b+R\theta\hat{\mathbf z}+\nabla^2\mathbf u,\qquad \dot{\mathbf b}=\partial_z\mathbf u+\zeta\nabla^2\mathbf b.}
$$
Complete the linear system with $\dot\theta=w+\nabla^2\theta$, $\nabla\cdot\mathbf u=0$ and the <solenoidal magnetic-field constraint> $\nabla\cdot\mathbf b=0$. The omitted advective and perturbed-field tension terms are quadratic in the disturbances. Magnetic tension resists bending of the vertical field, while diffusion allows that bending to relax.
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